SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Let $B^{\prime }{E}^{\prime }$, $C^{\prime }{F}^{\prime }$ be the altitudes of $△A{B}^{\prime }{C}^{\prime }$ and concur at $H$. We have that $A{B}^{\prime }P{C}^{\prime }$ is inscribed in the circle with diameter $AP$, so $AP$ passes through the circumcenter of $△A{B}^{\prime }{C}^{\prime }$. Then $AP$, $AH$ reflect each other by the angle bisector of $\angle B^{\prime }A{C}^{\prime }$. From that, easy to prove that $△AH{F}^{\prime }\sim △AP{C}^{\prime }$.



But $BE\parallel C^{\prime }{F}^{\prime }$, as Thales's Theorem, we have $\frac{AE}{A{F}^{\prime }}=\frac{AB}{A{C}^{\prime }}$, implies that $$\frac{HE}{PB}=\frac{AE}{AB}=\frac{EF}{BC}.$$ Similarly, we get $\frac{HF}{PC}=\frac{EF}{BC}$, thus $△HEF\sim △PBC$. Then we get $$\angle EHF=\angle BPC=90^{\circ }.$$ $\square$